AQA A-Level Chemistry: Chemical equilibria, Le Chatelier's principle and Kc (7405) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.1.6 (3.1.6.1 and 3.1.6.2) of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2 (July 2026), for AS and A-level exams from June 2016 onwards. All of it is AS content and also part of the full A-level; none of it is “A-level only”. They use different reactions and numbers from the study guide and the revision notes. Course links: AQA A-Level Chemistry hub | printable checklist.

Equilibrium constants below are practice values, not data-sheet values.

Questions

1. State two features of a reversible reaction at dynamic equilibrium. [2]

2. Nitrogen monoxide reacts with oxygen: 2NO(g) + O₂(g) ⇌ 2NO₂(g). Write an expression for Kc for this reaction and deduce its units. [2]

3. Methanol can be made by the reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g). The forward reaction is exothermic.

(a) State and explain the effect of increasing the total pressure on the equilibrium yield of methanol. [2] (b) State and explain the effect of increasing the temperature on the equilibrium yield of methanol. [2]

4. Explain the effect of adding a catalyst to a reaction mixture at equilibrium. Refer to the rates of reaction, the position of equilibrium and the value of Kc. [3]

5. For the equilibrium N₂O₄(g) ⇌ 2NO₂(g) at a fixed temperature, the equilibrium concentrations are [N₂O₄] = 0.0420 mol dm⁻³ and [NO₂] = 0.0130 mol dm⁻³. Calculate Kc and give its units. [3]

6. 0.600 mol of carbon monoxide and 0.600 mol of steam are sealed in a 2.00 dm³ container and heated. They reach equilibrium:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

At equilibrium, 0.400 mol of CO₂ is present.

(a) Calculate Kc for this reaction at this temperature. [4] (b) Explain why the value of Kc would be the same if the container volume had not been given. [1]

7. For the reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g) at a certain temperature, Kc = 12.5 mol⁻² dm⁶. An equilibrium mixture contains [CO] = 0.200 mol dm⁻³ and [H₂] = 0.400 mol dm⁻³. Calculate the equilibrium concentration of CH₃OH. [3]

8. For a homogeneous gas reaction, Kc = 160 at 400 K and Kc = 3.5 at 600 K.

(a) Deduce whether the forward reaction is exothermic or endothermic. Explain your answer. [2] (b) A student says: “Increasing the pressure on this reaction at 400 K will change the value of Kc.” Evaluate this statement. [2]

9. Ammonia is made in the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The forward reaction is exothermic and an iron catalyst is used.

(a) Explain why a compromise temperature is used rather than a low temperature. [3] (b) Explain why a compromise pressure is used rather than a very high pressure. [3] (c) Explain why the iron catalyst is used, even though it has no effect on the equilibrium yield. [2]

10. For CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, a student mixes 0.150 mol ethanoic acid, 0.150 mol ethanol and some dilute hydrochloric acid catalyst containing 0.0200 mol of water. The sealed mixture reaches equilibrium at constant temperature and is then titrated with 1.00 mol dm⁻³ sodium hydroxide: the titre is 55.00 cm³. A separate titration of the same volume of hydrochloric acid alone needs 5.00 cm³ of the same sodium hydroxide.

(a) Explain why the separate titration of hydrochloric acid alone is needed. [1] (b) Calculate the amount, in moles, of ethanoic acid in the equilibrium mixture. [2] (c) Calculate the amounts, in moles, of ethanol, ethyl ethanoate and water in the equilibrium mixture. [2] (d) Calculate Kc for this reaction. [2] (e) Explain why the total volume of the mixture is not needed to calculate Kc. [1]

11. 1.00 mol of hydrogen and 1.00 mol of iodine are heated in a sealed 2.00 dm³ flask. They reach equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g). At this temperature Kc = 64.0.

(a) Calculate the amount, in moles, of HI at equilibrium. [3] (b) Calculate the equilibrium concentration of H₂, and explain why you could find the amounts of each species without using the volume. [2] (c) More hydrogen is added at the same temperature. State and explain the effect on the value of Kc and on the amount of HI at equilibrium. [2]

12. For the equilibrium N₂O₄(g) ⇌ 2NO₂(g), the forward reaction is endothermic. For each change below, state the effect on the amount of NO₂ at equilibrium and on the value of Kc.

(i) temperature increased; (ii) total pressure increased at constant temperature; (iii) catalyst added; (iv) some NO₂ removed at constant temperature. [4]

Answers

1. The forward and reverse reactions proceed at equal rates [1]. The concentrations of reactants and products remain constant [1]. [2] Examiner insight: “Concentrations are equal” is a common wrong answer and scores nothing; the word needed is “constant”.

2. Kc = [NO₂]² / ([NO]²[O₂]) [1]. Units: (mol dm⁻³)² / (mol dm⁻³)³ = mol⁻¹ dm³ [1]. [2] Examiner insight: Square brackets are required in the expression; round brackets or missing powers lose the expression mark.

3. (a) Yield increases [1]: 3 moles of gas on the left and 1 on the right, so the equilibrium shifts to the side with fewer moles of gas to reduce the pressure [1]. (b) Yield decreases [1]: the equilibrium shifts in the endothermic direction, which is the reverse reaction, to absorb the added heat [1]. Examiner insight: Each two-mark part needs both the direction and a specific reason; “Le Chatelier’s principle” on its own does not earn the explanation mark.

4. A catalyst increases the rates of the forward and reverse reactions equally [1]. The position of equilibrium does not change, so the yield is unchanged [1]. Kc is unchanged, but equilibrium is reached faster [1]. [3] Examiner insight: Any statement that the catalyst shifts the equilibrium or increases the yield contradicts a correct point and can cancel the mark.

5. Kc = [NO₂]² / [N₂O₄] = 0.0130² / 0.0420 [1] = 4.02 × 10⁻³ [1] mol dm⁻³ [1]. [3] Examiner insight: The units mark is separate, so a correct number with wrong or missing units scores two of the three marks.

6. (a) Moles table: CO 0.600 − 0.400 = 0.200; H₂O 0.200; CO₂ 0.400; H₂ 0.400 mol [1]. Concentrations: [CO] = [H₂O] = 0.100, [CO₂] = [H₂] = 0.200 mol dm⁻³ [1]. Kc = [CO₂][H₂] / ([CO][H₂O]) = (0.200 × 0.200) / (0.100 × 0.100) [1] = 4.00 (no units) [1]. (b) There are 2 moles of gas on each side, so the volume terms cancel in the Kc expression [1]. Examiner insight: An error in the moles table carries forward; later marks can still be awarded for correct use of wrong values (error carried forward).

7. [CH₃OH] = Kc × [CO][H₂]² [1] = 12.5 × 0.200 × 0.400² [1] = 0.400 mol dm⁻³ [1]. [3] Examiner insight: Forgetting to square [H₂] gives 1.00 mol dm⁻³ and loses the last two marks; rearrange before substituting.

8. (a) Exothermic [1]: Kc decreases as temperature increases, so less product forms at higher temperature, meaning the equilibrium shifts left in the endothermic (reverse) direction [1]. (b) The statement is incorrect: Kc depends only on temperature [1]. A pressure change can only shift the position of equilibrium; Kc is unchanged [1]. Examiner insight: For “evaluate”, a bare “wrong” scores nothing; the mark comes from stating what Kc does depend on.

9. (a) A low temperature gives a higher equilibrium yield, because the forward reaction is exothermic [1]. But at a low temperature the rate is too slow [1]. A compromise temperature gives an acceptable yield at an acceptable rate [1]. (b) High pressure shifts the equilibrium right (2 moles of gas on the right, 4 on the left) and increases the rate [1]. Very high pressure needs expensive equipment and a lot of energy to compress the gases [1], and has safety risks, so the pressure is limited [1]. (c) The catalyst increases the rate, so equilibrium is reached faster [1]. A lower temperature can then be used, giving a better yield at a useful rate [1]. Examiner insight: Compromise answers need both sides of the trade-off; an answer about yield alone is capped at one mark per part.

10. (a) The hydrochloric acid catalyst also reacts with sodium hydroxide, so its titre must be subtracted to leave the titre due to ethanoic acid only [1]. (b) Titre for ethanoic acid = 55.00 − 5.00 = 50.00 cm³ [1]. Moles = 50.00/1000 × 1.00 = 0.0500 mol [1]. (c) Ethanoic acid reacted = 0.150 − 0.0500 = 0.100 mol, so ethanol = 0.0500 mol and ethyl ethanoate = 0.100 mol [1]. Water = 0.0200 + 0.100 = 0.120 mol [1]. (d) Kc = (0.100 × 0.120) / (0.0500 × 0.0500) [1] = 4.80 [1]. (e) There are 2 moles on each side, so the volume terms cancel [1]. Examiner insight: Missing the 0.0200 mol of water from the catalyst gives 4.00; this loses the water mark in (c) but the Kc mark in (d) can be awarded as error carried forward.

11. (a) Let x mol of H₂ react: H₂ = I₂ = 1.00 − x, HI = 2x; so (2x)² / (1.00 − x)² = 64.0 [1]. Square root: 2x / (1.00 − x) = 8.00, so 10x = 8.00 [1]. x = 0.800, so HI = 1.60 mol [1]. (b) [H₂] = (1.00 − 0.800) / 2.00 = 0.100 mol dm⁻³ [1]. There are 2 moles of gas on each side, so the volume cancels in the Kc expression [1]. (c) Kc is unchanged because the temperature is unchanged [1]. The equilibrium shifts right to use up some of the added hydrogen, so the amount of HI increases [1]. Examiner insight: Taking the square root keeps this a one-step solve; the other root from a quadratic (x = 1.33) needs more H₂ than was present and must be rejected.

12. (i) more NO₂; Kc increases [1]. (ii) less NO₂ (shifts to the side with fewer moles of gas); Kc unchanged [1]. (iii) no change in NO₂; Kc unchanged [1]. (iv) shifts right, so some NO₂ is replaced, but there is less NO₂ than before the removal; Kc unchanged [1]. [4] Examiner insight: Each mark needs both effects correct; only temperature changes Kc.

Where marks are usually lost

• Writing “equal concentrations” in a definition of dynamic equilibrium.
• Leaving out the specific reason (“fewer moles of gas”, “endothermic direction”) in a Le Chatelier explanation.
• Saying a catalyst increases the yield or shifts the equilibrium.
• Using equilibrium moles in Kc when the volume does not cancel (Questions 5 and 7 have unequal moles of gas).
• Forgetting water added with a catalyst solution, or not subtracting a blank titre.
• Missing or wrong units for Kc, or giving units when they cancel.
• Saying pressure or concentration changes alter Kc.
• Compromise answers that talk only about yield and ignore rate, cost or safety.

Next steps

• Revise weak areas with the revision notes.
• Reread the full explanations in the study guide.
• Practise the underlying mole calculations.
• See every topic on the AQA A-Level Chemistry hub and track progress on the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.1.6: Chemical equilibria, Le Chatelier’s principle and Kc.